Can you solve?

Question

1. -49x^2 + 28x = 4
2. 3x^4 + 4x^3 -9x^2 - 12x = 0
3. x^4 - 3x^2 + 4 = 0

Answer 1

1) 49x² - 28x + 4 = 0 can be factored

(7x - 2 ) ( 7x - 2 ) = 0 and x = 2/7 .

2) 3x⁴ + 4x³ - 9x² - 12x = 0 can be factored

3x² ( x² - 3 ) + 4x ( x² - 3 ) = 0 and again

x ( x² - 3 ) ( 3 x + 4 ) = 0 and x = 0 or -4/3 .

( Imaginary solutions : ± √3i )

3) x⁴ - 3x² + 4 = 0

There are no real solutions.

( Imaginary solutions : ± 4/3 ± i/2 ?)

Answer 2

1. Get everything on one side, and multiply thru by -1. Then 49 x^2 - 28 x + 4 = 0. This is a perfect square (7 x - 2)^2 = 0, so x = +/- sqrt(2/7).

3. Factor to (x^2 + 4)(x^2 -1) =0. Set each quadratic term equal to 0 and solve (you get 1,-1 and +2i and -2i)

2. If you group the even and odd-power terms terms with each other, you get (3x^2 [x^2-3] +4x
[x^2-3])= 0 Hafta run, you should be able to continue to a solution.
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Answer 3

huh???